Question

# A body of mass 2 kg is projected with a speed 4 m/s at an angle 45∘ to the horizontal. What will be its angular momentum about the point of projection when the body is at the highest point in its trajectory?

A
1.62 kg m2/s
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B
32.62 kg m2/s
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C
162 kg m2/s
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D
3.262 kg m2/s
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Solution

## The correct option is D 3.26√2 kg m2/s For given projectile motion, Range (R)=u2sin2θg=42sin(2×45∘)9.8=169.8 m Maximum height (Hmax)=u2sin2θ2g=49.8 m So at maximum height, →r=x ^i+y ^j=R2 ^i+Hmax ^j =(89.8 ^i+49.8 ^j) m and we know that at highest point velocity will be in x direction only, →u=ux ^i+uy ^j=u cosθ ^i+0 ^j =4 cos45∘ ^i+0 ^j=(4√2 ^i+0 ^j) m/s Now we know, Angular momentum about origin (→L)=→r×→p =→r×(m→u)=m(→r×→u)=2[(89.8 ^i+49.8 ^j)×(4√2 ^i+0 ^j)]=2×4×49.8×√2(^j×^i) =3.26√2(−^k) as, (^j×^i=−^k) |L|=3.26√2 kg m2/s Hence, option (d) is the correct answer. Alternate solution: We know that when the body is at the top most point: vy=0 , vx=vocosθ=4cos45∘=4√2m/s Angular momentum is given by: L=mass×velocity×perpendicular distance of velocity from the origin Here perpendicular distance is: Hmax=u2sin2θ2g=49.8 m Therefore, L=2×4√2×49.8 ⇒L=3.26√2kg m2/s

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